Convergence of a complex series on the boundary

Question

I have the following complex power-series: $$\sum_{n=1}^\infty \frac{z^{n!}}{n} $$ Its radius of convergence is $R=1$. I am trying to investigate its behavior on the boundary ($z$ such that $|z|=1$). It is quite easy to see that the series diverges at roots of unity so I am looking for a point on the boundary where it converges.

I tried $z=e^{e \pi i}$, which, by using Euler's identity, gives me: $$\sum_{n=1}^\infty \frac{z^{n!}}{n} = \sum_{n=1}^\infty \frac{e^{ie \pi n!}}{n}= \sum_{n=1}^\infty \frac{\cos(e \pi n!)}{n} + i \sum_{n=1}^\infty \frac{\sin(e \pi n!)}{n}. $$ I'm pretty sure that the first series on the right-hand side converges since $\sin(e \pi n!)$ should be infinitesimal of order 1 as $n \to \infty$ but I suspect that the second series diverges...

Any idea?

Answer 1

I don't know about a specific point where it converges, but in case it helps, the series converges at almost every boundary point.

Of course Carleson's theorem says that any $L^2$ Fourier series converges almost everywhere, but that's a stupendously deep result. The same result for lacunary series is much simpler. For example, the Fejer kernel is positive, hence has bounded $L^1$ norm, which implies that any $L^2$ Fourier series is Cesàro summable almost everywhere. For a lacunary series there's a tauberian theorem: Cesàro summability at a point implies convergence at that point.

In fact your series is so lacunary that getting convergence from Cesàro summability is very simple. If $n!<m<(n+1)!$ the obvious estimates imply that $$\lVert s_mf-\sigma_{(n+1)!}f\rVert_\infty^2\le\lVert f\rVert_2^2\sum_{k=1}^n\left|\left(\frac{k!}{(n+1)!}\right)^2\right|\to0\quad(n\to\infty).$$